Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, b) → f(a, c)
f(c, d) → f(b, d)
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, b) → f(a, c)
f(c, d) → f(b, d)
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(a, b) → f(a, c)
f(c, d) → f(b, d)
The signature Sigma is {f}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, b) → f(a, c)
f(c, d) → f(b, d)
The set Q consists of the following terms:
f(a, b)
f(c, d)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(a, b) → F(a, c)
F(c, d) → F(b, d)
The TRS R consists of the following rules:
f(a, b) → f(a, c)
f(c, d) → f(b, d)
The set Q consists of the following terms:
f(a, b)
f(c, d)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(a, b) → F(a, c)
F(c, d) → F(b, d)
The TRS R consists of the following rules:
f(a, b) → f(a, c)
f(c, d) → f(b, d)
The set Q consists of the following terms:
f(a, b)
f(c, d)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.